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A long straight wire, carrying a current...

A long straight wire, carrying a current of `200A`, runs through a cubical wooden box, entering and leaving through holes in the centres of opposite faces Figure. The length of each side of the box is `20cm`. Consider an element PQ of the wire of `1cm` long at the centre of the box. Calculate the magnitude of the magnetic field produced by this element at the points A, B, C and D as shown in figure. The points A, B and C are at the centres of the faces of the cube and the point D is at the mid point of one edge.

Text Solution

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Here, `I=200A`, `dl=1cm=0*01m`
For point A or B, `r=10cm=0*10m`, `theta=90^@`
Magnitude of the magnetic field at A or B due to current through PQ is
`=(mu_0)/(4pi)(Idlsintheta)/(r^2)=(10^-7xx200x(0*01))/((0*10)^2)sin90^@`
`=20xx10^-6T`
For point C, `theta=0^@` and `sin theta=0`
Magnitude of magnetic field at C due to current through PQ is
`=(mu_0)/(4pi)(Idlsin0^@)/(r^2)=0`
For Point D, distance from the element PQ will be
`r^'=sqrt(10^2+10^2)=10sqrt2cm=0*10sqrt2m`,
`theta=45^@`
Magnitude of magnetic field at D due to current through PQ is
`=(mu_0)/(4pi)(Idlsin45^@)/(r^('2))=10^-7xx(200xx(0*01)xx1//sqrt2)/((0*10)sqrt2)`
`=7*07xx10^-6T`
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